the sales tax in your city is 4.4 and an item costs $3 before tax how much do u pay on the item

Hi there! :)

Answer:

Subtract 43 from 20 raised to the 7th power; then multiply by 3. The answer is: 3,839,999,871

Step-by-step explanation:

"Subtract" means to take away a number from another.

" 20 raised to the 7th power" is the same thing as 20 exponent 7: [tex]20^{7}[/tex]

SO, you want to take away 43 from [tex]20^{7}[/tex] and then multiply the answer by 3:

([tex]20^{7}[/tex] - 43) × 3

[tex]20^{7} =[/tex] 1,280,000,000

(1,280,000,000 - 43) × 3

1,280,000,000 - 43 = 1,279,999,957

1,279,999,957 × 3

1,279,999,957 × 3 = 3,839,999,871

There you go! I really hope this helped, if there's anything just let me know! :)

Answer:

I got 3839999831

Step-by-step explanation:

So 20^7(calculator), then subtract and mult

Answer:

3(t+2)

Step-by-step explanation:

As with any problem involving division of fractions, you can invert the denominator and multiply.

Your knowledge of the factoring of the difference of squares helps. If that doesn't work for you, you can always use synthetic division or polynomial long division to find the quotient of (t^2-4) and (t-2).

[tex]\displaystyle\frac{\frac{4t^2-16}{8}}{\frac{t-2}{6}}=\frac{4(t^2-4)}{8}\cdot\frac{6}{t-2}\\\\=\frac{3(t+2)(t-2)}{t-2}=3(t+2)[/tex]

Answer:

Step-by-step explanation:

a) The mean of a normal distribution is also the median. Half the population will have values above the mean. Half of 200 is 100, so ...

... 100 students will have grades above 70%.

b) 84% is 14% above the mean. Each 7% is 1 standard deviation, so 14% is 2 standard deviations above the mean. The empirical rule tells you 95% of the population is within 2 standard deviations of the mean, so about 5% of students (10 students) got grades higher than 84% or lower than 56%. The normal distribution is symmetrical, so we expect about 5 students in each range.

... about 5 students will have grades above 84%.

1. The number of students who received a grade greater than 70% is about 100.

2. The number of students who got a grade higher than 84% is about 195.

To find the number of students who received a grade greater than 70%,

you need to calculate the area under the normal distribution curve to the right of 70%.

Similarly, to find the number of students who got a grade higher than 84%,

you need to calculate the area under the normal distribution curve to the right of 84%.

You can use the z-score formula to convert the given grades to z-scores, then look up the corresponding areas in the standard normal distribution table.

For a grade of 70%:

[tex]\[ Z = \frac{X - \mu}{\sigma}[/tex]

[tex]= \frac{70 - 70}{7}[/tex]

= 0

For a grade of 84%:

[tex]\[ Z = \frac{X - \mu}{\sigma}[/tex]

[tex]= \frac{84 - 70}{7}[/tex]

[tex]\approx 2[/tex]

Then, you look up the corresponding areas in the standard normal distribution table:

Now, to find the number of students:

Students with grades greater than 70%:

Number of students = Total students × Area to the right of 70%

= 200 × (1 - 0.5000)

= 200 × 0.5000

= 100

Students with grades higher than 84%:

Number of students = Total students × Area to the right of 84%

= 200 × (1 - 0.0228)

≈ 200 × 0.9772

≈ 195

So, the number of students who received a grade greater than 70% is about 100, and the number of students who got a grade higher than 84% is about 195.

Scientific notation has exactly one non-zero digit to the left of the decimal point in the mantissa. The numbers listed above do not.

_____

Comment on these numbers

There may be good reasons for writing the numbers in this form. For example, in engineering, it is often useful to have the exponent be a multiple of 3. In other instances, writing the number in these forms may facilitate arithmetic. However good the reasons may be, these numbers are not in in the form defined as "scientific notation."

∠BDC and ∠AED are right angles

Because ∠C ≅ ∠C, the additional bit of information above can be used to show AA similarity.

____

None of the other offered choices says anything about both triangles. In order to show similarity, you need information about corresponding parts of the two triangles. Information about one triangle alone is not sufficient.

Answer:

I think it's A. ∠BDC and ∠AED are right angles

Step-by-step explanation:

I hope this helps.

Answer:

72.25 in^2

Step-by-step explanation:

Formula

Area = s^2 where s is the length of a side.

Given

s = 8.5 in

Solution

Area = s^2

Area = 8.5^2 = 8.5 * 8.5

Area = 72.25 in^2

QUESTION 1a

The given polygon has vertices [tex]A(-2,3),B(3,1),C(-2,-1),D(-3,1)[/tex].

We plot the points and connect them to obtain the figure as shown in the attachment.

The polygon has four sides and two pairs of adjacent sides equal.

Therefore Polygon ABCD can be classified as a quadrilateral, specifically a kite.

QUESTION 1b

We can find the perimeter by adding the length of all the sides of the kite

[tex]Perimeter=|AB|+|BC|+|CD|+|AD|[/tex]

Or

[tex]Perimeter=2|AB|+2|AD|[/tex]

Recall the distance formula

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

We use the distance formula to find the length of each side.

[tex]|AB|=\sqrt{(3--2)^2+(1-3)^2}[/tex]

[tex]|AB|=\sqrt{(3+2)^2+(1-3)^2}[/tex]

[tex]|AB|=\sqrt{(5)^2+(-2)^2}[/tex]

[tex]|AB|=\sqrt{25+4}[/tex]

[tex]|AB|=\sqrt{29}=5.385[/tex]

Length of side AD

[tex]|AD|=\sqrt{(-3--2)^2+(1-3)^2}[/tex]

[tex]|AD|=\sqrt{(-3+2)^2+(1-3)^2}[/tex]

[tex]|AD|=\sqrt{(-1)^2+(-2)^2}[/tex]

[tex]|AD|=\sqrt{1+4}[/tex]

[tex]|AD|=\sqrt{5}=2.24[/tex]

[tex]Perimeter=2(5.1)+2(2.236)[/tex]

[tex]Perimeter=10.77+4.472[/tex]

[tex]Perimeter=15.242[/tex]

The perimeter of the kite to the nearest tenth is 15.2 units

QUESTION 1c

The area of kite ABCD is twice the area of ΔABD

[tex]Area\:of\:ABD=\frac{1}{2}\times |BD| \times |AE|[/tex]

[tex]Area\:of\:ABD=\frac{1}{2}\times 6 \times 2[/tex]

[tex]Area\:of\:ABD=6\:square\units[/tex]

Therefore the are of the kite is

[tex]=2\times 6=12[/tex]

The area of the kite is 12 square units.

QUESTION 2a

The vertices of the given polygon are

[tex]P(-3,-4),Q(3,-3),R(3,-2),S(-3,2)[/tex].

We plot all the four points as shown in the diagram in the attachment.

The polygon has one pair of opposite sides parallel and has four sides.

The polygon is a quadrilateral, specifically a trap-ezoid.

QUESTION 2b

The area of the trap-ezoid can be found using the formula

[tex]Area=\frac{1}{2}(|RQ|+|PS|)\times |RU|[/tex].

We use the absolute value method to find the length of RQ,PS and RU because they are vertical and horizontal lines.

[tex]|RQ|=|-3--2|[/tex]

[tex]|RQ|=|-3+2|[/tex]

[tex]|RQ|=|-1|[/tex]

[tex]|RQ|=1[/tex]

The length of PS is

[tex]|PS|=|-4-2|[/tex]

[tex]|PS|=|-6|[/tex]

[tex]|PS|=6[/tex]

The length of RU

[tex]|RU|=|-3-3|[/tex]

[tex]|RU|=|-6|[/tex]

[tex]|RU|=6[/tex]

The area of the trap-ezoid is

[tex]Area=\frac{1}{2}(1+6)\times 6[/tex].

[tex]Area=(7)\times 3[/tex].

[tex]Area=21[/tex].

Therefore the area of the trap-ezoid is 21 square units.

QUESTION 2c

The perimeter of the trap-ezoid

[tex]=|PQ|+|RS|+|QR|+|PS|[/tex]

We use the distance formula to determine the length of RS and PQ.

[tex]|RS|=\sqrt{(3--3)^2+(-2-2)^2}[/tex]

[tex]|RS|=\sqrt{(3+3)^2+(-2-2)^2}[/tex]

[tex]|RS|=\sqrt{(6)^2+(-4)^2}[/tex]

[tex]|RS|=\sqrt{36+16}[/tex]

[tex]|RS|=\sqrt{52}[/tex]

[tex]|RS|=7.211[/tex]

We now calculate the length of PQ

[tex]|PQ|=\sqrt{(3--3)^2+(-3--4)^2}[/tex]

[tex]|PQ|=\sqrt{(3+3)^2+(-3+4)^2}[/tex]

[tex]|PQ|=\sqrt{(6)^2+(1)^2}[/tex]

[tex]|PQ|=\sqrt{36+1}[/tex]

[tex]|PQ|=\sqrt{37}[/tex]

[tex]|PQ|=6.083[/tex]

We already found that,

[tex]|PS|=6[/tex]

and

[tex]|RQ|=1[/tex]

We substitute all these values to get,

[tex]Perimeter=6+1+7.211+6.083[/tex]

[tex]Perimeter=20.294[/tex]

To the nearest tenth, the perimeter quadrilateral PQRS is 20.3 units.

QUESTION 3a

The given polygon has vertices

[tex]E(-4,1),F(-2,3),G(-2,-4)[/tex]

We plot all the three points to the polygon shown in the diagram. See attachment.

The polygon has three unequal sides, therefore it is a triangle, specifically scalene triangle.

QUESTION 3b

We can calculate the area of this triangle using the formula,

[tex]Area=\frac{1}{2} \times |FG| \times |EH|[/tex] see attachment

We can use the absolute value method to find the length of FG and EH because they are vertical or horizontal lines.

[tex]|FG|=|-4-3|[/tex]

[tex]|FG|=|-7|[/tex]

[tex]|FG|=7[/tex]

Now the length of EH is

[tex]|EH|=|-4--2|[/tex]

[tex]|EH|=|-4+2|[/tex]

[tex]|EH|=|-2|[/tex]

[tex]|EH|=2[/tex]

The area is

[tex]Area=\frac{1}{2} \times 7 \times 2[/tex]

[tex]Area=7[/tex]

Therefore the area of the triangle is 7 square units.

QUESTION 3c

The perimeter of the triangle can be found by adding the length of the three sides of the triangle.

[tex]Perimeter=|EF|+|FG|+|GE|[/tex]

The length of EF can be found using the distance formula,

[tex]|EF|=\sqrt{(-2--4)^2+(3-1)^2}[/tex]

[tex]|EF|=\sqrt{(-2+4)^2+(3-1)^2}[/tex]

[tex]|EF|=\sqrt{(2)^2+(2)^2}[/tex]

[tex]|EF|=\sqrt{4+4}[/tex]

[tex]|EF|=\sqrt{8}[/tex]

[tex]|EF|=2.828[/tex]

The length of EG can also be found using the distance formula

[tex]|EG|=\sqrt{(-4--2)^2+(1--4)^2}[/tex]

[tex]|EG|=\sqrt{(-4+2)^2+(1+4)^2}[/tex]

[tex]|EG|=\sqrt{(-2)^2+(5)^2}[/tex]

[tex]|EG|=\sqrt{4+25}[/tex]

[tex]|EG|=\sqrt{29}[/tex]

[tex]|EG|=5.385}[/tex]

We found [tex]|FG|=7[/tex]

The perimeter of the triangle is

[tex]Perimeter=5.385+7+2.828[/tex]

[tex]Perimeter=15.213[/tex]

Therefore the perimeter of the triangle is 15.2 units to the nearest tenth

DE = 1.5 cm, EF = 2.25 cm, FD = 3 cm

Side DE corresponds to side AB and is (1.5 cm)/(2 cm) = 3/4 the length of it.

Because the triangles are similar, all pairs of corresponding sides have the same ratio. That is, the side lengths in ΔDEF are 3/4 of those in ΔABC.

... EF = (3/4)×BC = (3/4)×(3 cm)

... EF = 2.25 cm

... FD = (3/4)×CA = (3/4)×(4 cm)

... FD = 3 cm

The lengths of the sides of triangle △DEF, given that AB=2 cm, BC=3 cm, CA=4 cm, and DE=1.5 cm, would be EF = 2.25 cm and DF = 3 cm.

If triangles △ABC and △DEF are similar, then the ratio of the corresponding sides is the same. We know AB = 2 cm which corresponds to DE = 1.5 cm in the triangle △DEF. The ratio of AB to DE is 2cm:1.5cm or 4:3. Let's compute the sides of the triangle △DEF using this ratio:

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Assuming your target boxes are (-12x +36) and (8x -24), here are the classifications of the expressions with parentheses.

-3(4x + 12) = -12x -36 (neither)

-4(3x -9) = -12x +36

4(2x - 6) = 8x -24

-4(-2x - 6) = 8x +24 (neither)

3(-4x +12) = -12x +36

2(4x - 12) = 8x -24

_____

The distributive property applies:

... a(b+c) = ab +ac

Same signs multiply to give positive. Different signs multiply to give negative.

Expressions like -3(4x + 12), -4(3x -9), and 3(-4x +12) are equivalent to -12x + 36 while 4(2x - 6) and 2(4x - 12) are equivalent to 8x - 24. The expression -4(-2x -6) is not equivalent to others.

The question is asking to categorize the expressions whether they are equivalent or not. Two expressions are equivalent if they have the same value for all values of their variables. Let's evaluate each expression:

So we see that -3(4x + 12), -4(3x -9), and 3(-4x +12) are equivalent to -12x + 36, while 4(2x - 6) and 2(4x - 12) are equivalent to 8x - 24. The expression -4(-2x -6) doesn't match any other expression.

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There is an infinite number of numbers between 1 and 1.01 on the number line. The correct answer should theoretically be: 1 < n < 1.01.

Answer:

Length of wire = 50 ft

Step-by-step explanation:

In the figure below AB is the telephone pole and C is the point on the ground where wire is anchored.

so we have

AB= 45 ft

∠ACB= 64°

Let is assume the length of the wire be x ft

now in Δ ABC

[tex]sin C = \frac{AB}{AC}[/tex]

we can plug AB= 45, ∠C= 64°and AC=x

so we have

[tex]sin(64)[/tex]°[tex]=\frac{45}{x}[/tex]

[tex]xsin(64)[/tex]°[tex]=45[/tex]

[tex]x=\frac{45}{sin(64)}[/tex]

now we have sin 64°=0.8988

[tex]x=\frac{45}{0.8988}[/tex]

[tex]x=50.07[/tex] ft

rounding to nearest foot

x= 50 ft

Final answer:

To find the length of the wire attached to a telephone pole at a 64-degree angle, we use the tangent function, resulting in an approximate wire length of 22 feet to the nearest foot.

Explanation:

The question asks how long a wire is if it's attached to the top of a 45-foot telephone pole and anchored to the ground, with an angle of elevation of 64 degrees.

To solve this, we can use trigonometry, specifically the tangent function, because we know the opposite side (height of the pole) and are trying to find the hypotenuse (length of the wire).

The tangent of an angle in a right triangle is the opposite side divided by the adjacent side. In this scenario, the angle of elevation is 64 degrees and the opposite side (height of the telephone pole) is 45 feet.

To find the length of the wire (hypotenuse), we use the formula: length of wire = opposite side / tan(angle), which translates to length of wire = 45 / tan(64 degrees).

Using a calculator, we find that tan(64 degrees) is approximately 2.05.

Therefore, the length of the wire is approximately 45 / 2.05, which equals about 21.95 feet.

To the nearest foot, the wire would be 22 feet long.

Answer:

B. -2

Step-by-step explanation:

We have been given an expression [tex]\frac{P}{3Q}[/tex] and we are asked to evaluate our expression, when P=-24 and Q=4.

Let us substitute P=-24 and Q=4 in our given expression.

[tex]\frac{-24}{3*4}[/tex]

Let us multiply 3 by 4.

[tex]\frac{-24}{12}[/tex]

[tex]-\frac{24}{12}[/tex]

Upon dividing 24 by 12 we will get,

[tex]-\frac{24}{12}=-2[/tex]

Therefore our expression simplifies to -2 and option B is the correct choice.

(2π-5) radians ≈ 1.2831853 radians

The reference angle is the magnitude of the acute angle made with the x-axis. The value of θ is greater than 3π/2, so the reference angle is ...

... θ' = abs(θ -2π) = 2π-5 . . . . radians

$67.45

There are a couple of ways you can go at these.

1. Compute the tip, then add that to the bill.

... 15% × $58.65 = $8.7975 ≈ $8.80

... $58.65 + 8.80 = $67.45

2. Compute the effect of adding the tip, then use that result to multiply by the bill amount.

... bill + (15% × bill) = bill × (1 +0.15) = 1.15 × bill

... 1.15 × $58.65 = $67.4475 ≈ $67.45

_____

Comment on percentages

The symbol % is a fancy (shorthand) way to write /100. The terminology "per cent" means "per hundred" or "divided by 100".

So, 15% = 15/100.

From your knowledge of place value and the meaning of decimal numbers, you know 15/100 (fifteen hundredths) = 0.15 (fifteen hundredths).

_____

Comment on 15% tip

10% = 10/100 = 1/10 of something can be computed by moving the decimal point one place left (dividing by 10).

5% of something is half of 10% of it.

15% is the sum of 10% and 5%.

Your bill is 58.65, so 10% of the bill is 5.865 ≈ 5.87. Half that is 2.9325 ≈ 2.93. The sum of 10% and 5% will be 5.87 +2.93 = 8.80, the amount that is 15% of the bill. Often, the numbers are such that this arithmetic can be done in your head.

Answer:

When dealing with VOLUME, an increase in a linear quantity, produces a third power result in the volume.

Increase the sides of a cube by 2 produces an 8 times effect in the volume.

Increasing each side of a cube by 4 produces a chnage of 4 * 4 * 4 or

64 times in the volume.

Step-by-step explanation:

Increasing the length of each side of a cube by a factor of 4 results in the volume increasing by a factor of 64, since volume is proportional to the cube of the linear dimensions.

When the length of each side of a cube is increased by a factor of 4, the effect on the volume of the cube is that it increases by a factor of [tex]4^3[/tex], or 64. This is because volume is a three-dimensional measure, and when each dimension (length, width, height) of a cube is multiplied by a factor, the volume is multiplied by the factor raised to the third power (since volume is calculated by length  *width * height).

Therefore, if the original length of one side of the cube is L, the original volume is L3. After increasing each side by a factor of 4, the new length becomes 4L, making the new volume [tex](4L)^3 = 4^3 \times L^3 = 64L^3[/tex]. Hence, the new volume is 64 times the original volume, not simply 4 times because the increase happens in each of the three dimensions.

Answer:

The graph shows that typically students with more hours of sleep(independent) had higher test scores (dependent)

Step-by-step explanation:

C. 75°

The sum of the three angles is 180°, as it is for the angles in any triangle.

35° + 70° + x° = 180° . . . . an equation expressing the relationship of the angles

x° = 180° -35° -70° . . . . . subtract 35° and 70°

x° = 75°

Answer:

120 cm^3

Step-by-step explanation:

The surface areas are in the ratio 60 to 135  so the single dimensions are in the ratio √60 to √135.

Therefore the volumes are in the ratio (√60)^3 to  (√135)^3  or 60^3/2 to  135^3/2.

So  Volume of  Pyramid B / Volume of Pyramid A

= 60^3/2 / 135^3/2.

Therefore we have the equation  60^3/2 / 135^3/2 =  V / 405  where V is the volume of pyramid B.

V = (60^3/2 * 405) / 135^3/2

=  120  cm^3

Answer:

120cm³

Step-by-step explanation:

√135/√60=3/2

3/2=1.5

1.5³=3.375

405/3.375=120

120cm³

Answer:

Add 78 to 54 to get 132 total patrons, then divide the number of females into the total number of patrons. Therefore, 78/132 = .5909 = 59%

Step-by-step explanation:

Answer:

Thats simple!

Step-by-step explanation:

G.o.o.g.l.e.

Answer:

The unit rate is [tex]\frac{2}{5}[/tex] km per minute. It means sarah bikes [tex]\frac{2}{5}[/tex] km in 1 minute or the speed of sarah is [tex]\frac{2}{5}[/tex] km per minute.

Step-by-step explanation:

It is given that Sarah bikes [tex]\frac{4}{5}[/tex] km in 2 min. It means Sarah covers [tex]\frac{4}{5}[/tex] km in 2 min.

The units rate is defined as

[tex]\text{Unit rate}=\frac{\text{Distance covered}}{\text{Time Taken}}[/tex]

[tex]\text{Unit rate}=\frac{(\frac{4}{5})}{2}[/tex]

[tex]\text{Unit rate}=\frac{4}{5}\times \frac{1}{2}[/tex]

[tex]\text{Unit rate}=\frac{2}{5}[/tex]

Therefore the unit rate is [tex]\frac{2}{5}[/tex] km per minute. It means sarah bikes [tex]\frac{2}{5}[/tex] km in 1 minute or the speed of sarah is [tex]\frac{2}{5}[/tex] km per minute.

Answer:

[tex]\ast=16a^2[/tex]

Step-by-step explanation:

Use formula for  a square of a sum:

[tex](x+y)^2=x^2+2xy+y^2.[/tex]

Note that

Then instead of * should be the square of the term 4a that is [tex](4a)^2=16a^2.[/tex]

Then

[tex]16a^2+56a+49=(4a+7)^2.[/tex]

To complete the pattern in the expression ∗ +56a+49 to make it a square of a sum or a difference, we can use the formula for (a+b)² as a guide. If 49 is b² and 56a is 2ab, the missing part would be a². On solving, we find that the value of a² is 16a².

The subject of this question is finding a monomial such that the given expression can be rewritten as a square of a sum or a difference. The expression is ∗ +56a+49. We can think of the square of a sum or a difference as the result of the formula (a+b)² = a² + 2ab + b² which gives a hint that we can structure our expression in a similar way.

Let's rearrange the given expression a bit using this hint. If we look at 49, it is equal to 7². This could be our b². Additionally, 56a can be written as 2*7*a or 2ab. So the missing part would be a² which completes the pattern of the formula.

To find the value of a², we need to know the value of a. Here's where 56a comes into play. It is 2ab (or in this case, 2*7*a). Solving this equation for a, we get a=4. So, a²=16. Therefore, the monomial that completes the pattern in the expression is 16a².

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Answer:

A. x^2 = y^2/3 -2

Step-by-step explanation:

Add 3x^2 to both sides of the equation. We do this so that the coefficient of x^2 is positive. It can be less confusing that way.

... y^2 = 3x^2 +6

Divide by 3.

... y^2/3 = x^2 +2

Subtract 2.

... y^2/3 -2 = x^2 . . . . . matches selection A.

Answer:

A. x^2 = y^2/3 -2

Step-by-step explanation:

A)

B) It is a matter of definition: speed = distance/time. Speed is proportional to distance and inversely proportional to time.

Speed is defined as the ratio of distance to the time required to cover that distance:

... speed = distance/time

Solving this relation for time, we have ...

... time = distance/speed

a) For each of the modes of transportation, we can find the time by using this relation.

... motorcyle: (150 km)/(75 km/h) = 2 h

... bus: (150 km)/(60 km/h) = 2.5 h

... truck: (150 km)/(50 km/h) = 3 h

... bicycle: (150 km)/(20 km/h) = 7.5 h

b) For a given distance, speed and time are inversely related as a matter of definition.

D. There are no solutions

First inequality:

... -16 ≥ 8x . . . . . . . add 8x-60 to the inequality

... -2 ≥ x . . . . . . . . divide by 8

Second inequality:

... -8 < 4x . . . . . . . . add 4x-58 to the inequality

... -2 < x . . . . . . . . . divide by 4

The problem statement requires a solution satisfy both conditions on x. The two solution sets are disjoint (have no points in common), so ...

... there are no solutions.

Answer:

4x-2=0

5x-3=0

Step-by-step explanation:

Answer:

31.25 %

Step-by-step explanation:

The fraction bar also means divide:

5 / 16

=

5 Divided By 16

=

0.3125

0.3125 * 100

= 31.25

- I.A -

The fraction 5/16 is converted to a percent by dividing 5 by 16 and then multiplying the result by 100, which equals 31.25%.

To convert the fraction 5/16 into a percent, we want to know how many parts out of 100 this fraction represents.

A percent is a way of expressing a fractional amount as a part of 100.

To do this, we can set up a proportion or multiply the fraction by 100.

So, we take 5 divided by 16 and then multiply the result by 100.

Doing the math:

= (5 / 16) x 100

= 0.3125 x 100

= 31.25%.

Therefore, the fraction 5/16 expressed as a percent is 31.25%.

B. MN/LN = RS/QR

ΔQRS ~ ΔLMN so corresponding segments are ...

Corresponding segments have the same ratio. So, ...

... LM/LN = QR/QS . . . . does not match A

... MN/LM = RS/QR . . . . matches B

... LM/MN = QR/RS . . . . does not match C

... MN/LM = RS/QR . . . . does not match D